Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 139 / 2 / vi / b / Solution

The strict statement as printed needs the qualification if $F\ne 0$. Part (a) already shows $F^2=0$ in the present situation, and the next part proves $F=0$; an unconditional $F^2<0$ would contradict that conclusion.

To prove the needed conditional statement, suppose $F\ne 0$. A fixed part has $h^0(X,F)=1$. Indeed, a different $F^{\prime }\in |F|$ has smaller multiplicity along some component of $F$; otherwise $F^{\prime }-F$ would be a nonzero effective linearly trivial divisor. Adding a member of the movable part avoiding that component contradicts fixedness.

The Hodge index theorem for algebraic surfaces, together with $D\cdot F=0$ and $D^2=0$, gives $F^2\le 0$. If equality held, Riemann–Roch theorem for algebraic surfaces would give $\chi (X,F)=2$, and Serre duality would give $h^2(X,F)=h^0(X,-F)=0$. Hence $h^0(X,F)=2+h^1(X,F)\ge 2$, a contradiction. ThereforeThis is the fixed-part elimination on a K3 surface argument.